## Tuesday, June 9, 2015

### What’s that Lambda?

To understand lambda ($\lambda$), you have to understand the producer’s problem and it optimum solution. You can find information (here). I stress in that page that we will use equation ${{X}_{ij}}=\frac{{{\alpha }_{ij}}\lambda {{X}_{j}}}{{{P}_{i}}}$  frequently.
\begin{align} & {{X}_{ij}}=\frac{{{\alpha }_{ij}}\lambda {{X}_{j}}}{{{P}_{i}}} \\ & or,{{P}_{i}}{{X}_{ij}}={{\alpha }_{ij}}\lambda {{X}_{j}} \\ \end{align}
Taking summation, we get
$\sum{{{P}_{i}}{{X}_{ij}}}=\sum{{{\alpha }_{ij}}\lambda {{X}_{j}}}$
Now, let’s view $\sum{{{P}_{i}}{{X}_{ij}}}$ as cost, so $\sum{{{P}_{i}}{{X}_{ij}}}={{C}_{j}}$ .

We also assume a Zero Profit Condition (ZPC) on CGE model. ZPC implies the cost of production of each commodity $\sum{{{P}_{i}}{{X}_{ij}}}$ equates to value of gross output $\sum{{{P}_{i}}{{X}_{j}}}$ so, we can write, $\sum{{{P}_{i}}{{X}_{ij}}}={{C}_{j}}=\sum{{{P}_{i}}{{X}_{j}}}$. Since,$\sum{{{P}_{i}}{{X}_{ij}}}=\sum{{{\alpha }_{ij}}\lambda {{X}_{j}}}={{C}_{j}}=\sum{{{P}_{i}}{{X}_{j}}}$.

Can you find the cell that represents above notation in this IO table?

Let’s consider
\begin{align} & \sum{{{P}_{i}}{{X}_{j}}}=\sum{{{\alpha }_{ij}}\lambda {{X}_{j}}} \\ & or,\sum{{{P}_{i}}{{X}_{j}}}=\sum{1\lambda {{X}_{j}}} \\ & \because \sum{{{\alpha }_{ij}}}=1 \\ & or,{{P}_{i}}{{X}_{j}}=\lambda {{X}_{j}} \\ & \therefore {{P}_{i}}=\lambda \\ \end{align}
From, the previous post (here), we know that,
$\lambda =Q\prod\limits_{i=1}^{4}{P_{i}^{{{\alpha }_{ij}}}}$
And $Q=\left( \frac{1}{{{A}_{j}}}\prod\limits_{i=1}^{4}{\alpha _{ij}^{-{{\alpha }_{ij}}}} \right)$
$\therefore {{P}_{i}}=\lambda =Q\prod\limits_{i=1}^{4}{P_{i}^{{{\alpha }_{ij}}}}$

### Let’s Linearize now

For that let’s take log on both sides
\begin{align} & Ln({{P}_{i}})=Ln\left( Q\prod\limits_{i=1}^{4}{P_{i}^{{{\alpha }_{ij}}}} \right) \\ & or,Ln({{P}_{i}})=Ln\left( QP_{1}^{{{\alpha }_{1j}}}P_{2}^{{{\alpha }_{2j}}}P_{3}^{{{\alpha }_{3j}}}P_{4}^{{{\alpha }_{4j}}} \right) \\ & or,Ln({{P}_{i}})=Ln(Q)+Ln(P_{1}^{{{\alpha }_{1j}}})+Ln(P_{2}^{{{\alpha }_{2j}}})+Ln(P_{3}^{{{\alpha }_{3j}}})+Ln(P_{4}^{{{\alpha }_{4j}}}) \\ & or,Ln({{P}_{i}})=Ln(Q)+{{\alpha }_{1j}}Ln({{P}_{1}})+{{\alpha }_{2j}}Ln({{P}_{2}})+{{\alpha }_{3j}}Ln({{P}_{3}})+{{\alpha }_{4j}}Ln({{P}_{4}}) \\ & or,Ln({{P}_{i}})=Ln(Q)+\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}Ln({{P}_{i}})} \\ \end{align}
Now, differentiate w.r.t ${{P}_{i}}$ ,
\begin{align} & or,\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{P}_{i}}}=\frac{1}{Q}\frac{dQ}{d{{P}_{i}}}+\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{P}_{i}}}} \\ & here, \\ & \frac{dQ}{d{{P}_{i}}}=0 \\ & \frac{d{{P}_{i}}}{d{{P}_{i}}}=1 \\ & \frac{1}{{{P}_{i}}}=0+\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{P}_{i}}}} \\ \end{align}
Taking $\frac{1}{d{{P}_{i}}}$ common,
\begin{align} & or,\frac{1}{{{P}_{i}}}=\frac{1}{d{{P}_{i}}}\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{d{{P}_{i}}}{{{P}_{i}}}} \\ & or,\frac{d{{P}_{i}}}{{{P}_{i}}}=\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{d{{P}_{i}}}{{{P}_{i}}}} \\ \end{align}
Here, $\frac{d{{P}_{i}}}{{{P}_{i}}}$ is the change in price ${{P}_{i}}$ , Lets denote by ${{p}_{i}}$ , so the linearized equation will be

$\therefore {{p}_{i}}=\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}{{p}_{i}}}$